Question

Proving Identities

Verify the identity$\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\mathbf{=}\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }\mathbf{+}\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }$

Short answer

The expression$\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\mathbf{=}\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }\mathbf{+}\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }$ is an identity.

Step-by-step solution

Step 1. Given information

An expression$\frac{\cos \theta }{1-\sin \theta }=\sec \theta +\tan \theta$

Step 2. Concept used

To see if the phrase has an identity, split it into two parts, LHS and RHS. Make both of them simpler after that. The phrase is considered to be an identity if the results are the same.

Step 3. Calculation

Now, simplify LHS of $\frac{\cos \theta }{1-\sin \theta }=\sec \theta +\tan \theta$

$\begin{array}{l}=\frac{\cos \theta }{1-\sin \theta }\\ =\frac{\cos \theta }{1-\sin \theta }\times \frac{1+\sin \theta }{1+\sin \theta }\\ =\frac{\cos \theta (1+\sin \theta )}{\left(1-{\sin }^2\theta \right)}\\ =\frac{\cos \theta (1+\sin \theta )}{{\cos }^2\theta }\\ =\frac{(1+\sin \theta )}{\cos \theta }\\ =\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }\\ =\sec \theta +\tan \theta \end{array}$

RHS of the equation is$\sec \theta +\tan \theta$

Both RHS and LHS are equal. Hence it is an identity